shhh... A Big Secret...No Wave Overhead, Just In Your Face.

From previous episodes of Amnesia's Dream,

$\cfrac { \partial T }{ \partial t } =\cfrac { x }{ g_{ T } } \cfrac { \partial ^{ 3 }T }{ \partial ^{ 3 }t }$

and so,

$\cfrac { dT(t) }{ dt } T(x)=\cfrac { x }{ g_{ T } } \cfrac { \partial  }{ \partial t } (\cfrac { \partial ^{ 2 }T }{ \partial t^{ 2 } } )$

From the wave equation,

$\cfrac { dT(t) }{ dt } T(x)=\cfrac { x }{ g_{ T } } \cfrac { \partial  }{ \partial t } (v^{ 2 }\cfrac { \partial ^{ 2 }T }{ \partial x^{ 2 } } )$

$\cfrac { dT(t) }{ dt } T(x)=\cfrac { x }{ g_{ T } } \left\{ 2v\cfrac { dv }{ dt } \cfrac { \partial ^{ 2 }T }{ \partial x^{ 2 } } +v^{ 2 }\cfrac { \partial ^{ 2 } }{ \partial x^{ 2 } } (\cfrac { \partial T }{ \partial t } ) \right\}$

If we assume that acceleration is wholly due to $g_T$,

$\cfrac{dv}{dt}=g_T$

$\cfrac { dT(t) }{ dt } T(x)=\cfrac { x }{ g_{ T } } \left\{ 2vg_{ T }T(t)\cfrac { d^{ 2 }T(x) }{ dx^{ 2 } } +v^{ 2 }\cfrac { \partial ^{ 2 } }{ \partial x^{ 2 } } (T(x)\cfrac { dT(t) }{ dt } ) \right\}$

$\cfrac { dT(t) }{ dt } T(x)=\cfrac { x }{ g_{ T } } \left\{ 2vg_{ T }T(t)\cfrac { d^{ 2 }T(x) }{ dx^{ 2 } } +v^{ 2 }\cfrac { dT(t) }{ dt } \cfrac { d^{ 2 }T(x) }{ dx^{ 2 } }  \right\}$

$\cfrac { dT(t) }{ dt } T(x)=2xvT(t)\cfrac { d^{ 2 }T(x) }{ dx^{ 2 } } +\cfrac { dT(t) }{ dt } v^{ 2 }\cfrac { x }{ g_{ T } } \cfrac { d^{ 2 }T(x) }{ dx^{ 2 } }$

Compare this with,

$T(x)= v^{ 2 }\cfrac { x }{ g_{ T } } \cfrac { d^{ 2 }T(x) }{ dx^{ 2 } }$

this suggests,

$2xvT(t)\cfrac { d^{ 2 }T(x) }{ dx^{ 2 } } =2xv\cfrac { \partial ^{ 2 }T }{ \partial x^{ 2 } } =\cfrac { 2x }{ v } \cfrac { \partial ^{ 2 }T }{ \partial t^{ 2 } } =0$

which might mean,

$\cfrac { \partial^{ 2 }T }{ \partial x^{ 2 } } =\cfrac { \partial ^{ 2 }T }{ \partial t^{ 2 } }=0$

then there is no wave in the radial direction, not of the form $T=T(t)T(x)$.  Just attenuation.